2005 AIME I Problems/Problem 2

Problem

For each positive integer $k$ , let $S_k$ denote the increasing arithmetic sequence of integers whose first term is 1 and whose common difference is $k$ . For example, $S_3$ is the sequence $1,4,7,10,\ldots.$ For how many values of $k$ does $S_k$ contain the term 2005?

Solution

Suppose that the $n$ th term of the sequence $S_k$ is 2005. Then $1+(n-1)k=2005$ so $k(n-1)=2004=2^2\cdot 3\cdot 167$ . The ordered pairs $(k,n-1)$ of positive integers that satisfy the last equation are $(1,2004)$ , $(2,1002)$ , $(3,668)$ , $(4,501)$ , $(6,334)$ , $(12,167)$ , $(167,12)$ , $(334,6)$ , $(501,4)$ , $(668,3)$ , $(1002,2)$ and $(2004,1)$ , and each of these gives a possible value of $k$ . Thus the requested number of values is 12.

Alternatively, notice that the formula for the number of divisors states that there are $(2 + 1)(1 + 1)(1 + 1) = 12$ divisors of $2^23^1167^1$ .

2005 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search

2005 AIME I Problems/Problem 2

Problem

Solution

See also